Find the critical points for the function g(x) = x^2\ln(3x) + 91.

Question:

Find the critical points for the function {eq}g(x) = x^2\ln(3x) + 91. {/eq}

Critical Point:

Generally, we can determine the critical points of a defined function using differentiation at which it is zero or undefined. Using the general rules and formula of derivatives we can easily take the derivative of the given function. Example of this is the sum rule, which expressed as {eq}\left(f\pm g\right)'=f'\pm g' {/eq}.

Answer and Explanation:

We can easily find the critical points of the given function by getting its derivative. Given the function, we have

{eq}\begin{align} g(x) &= x^2\ln(3x) + 91 \\ g'(x)&=\frac{d}{dx}\left(x^2\ln \left(3x\right)+91\right) \\ &= \frac{d}{dx}\left(x^2\ln \left(3x\right)\right)+\frac{d}{dx}\left(91\right) \end{align} {/eq}

For the first term, we need to perform the product rule, defined as

{eq}\left(f\cdot g\right)'=f'\cdot g+f\cdot g' {/eq}

We let {eq}f=x^2 {/eq} and {eq}g=\ln \left(3x\right) {/eq}, then we have

{eq}\begin{align} &= \left(\frac{d}{dx}\left(x^2\right)\ln \left(3x\right)+\frac{d}{dx}\left(\ln \left(3x\right)\right)x^2\right)+ \frac{d}{dx}\left(91\right)\\ &= \left(\left(2x\ln \left(3x\right)\right) + \frac{d}{dx}\left(\ln \left(3x\right)\right)x^2\right)+0\\ \end{align} {/eq}

Notice that we need to apply the chain rule to evaluate the derivative in the second term. We let {eq}f=\ln\left(u\right) {/eq} and {eq}u=3x {/eq}. Then, the chain rule is defined as

{eq}\begin{align} \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \end{align} {/eq},

{eq}\begin{align} &= \left(\left(2x\ln \left(3x\right)\right) + \frac{d}{du}\left(\ln \left(u\right)\right)\frac{d}{dx}\left(3x\right)x^2\right)+0\\ &= 2x\ln \left(3x\right)+\left(\frac{1}{u}\right)\left( 3\right)x^2 \\ &= 2x\ln \left(3x\right)+\left(\frac{1}{3x}\right)\left( 3\right)x^2 \\ &= 2x\ln \left(3x\right)+\frac{1}{x}x^2 \\ &= 2x\ln \left(3x\right)+x \end{align} {/eq}

Simplifying,

{eq}\begin{align} 2x\ln \left(3x\right)+x&=0\\ x\left(2\ln \left(3x\right)+1\right)&=0 \end{align} {/eq}

Now, to get the critical point, we need to use the zero factor principle. Then, we have

{eq}\begin{align} 2\ln \left(3x\right)+1&=0\\ 2\ln \left(3x\right)&=-1 \\ \ln \left(3x\right)&=-\frac{1}{2}\\ \end{align} {/eq}

To solve for x, recall the log rule {eq}a=\log _b\left(b^a\right) {/eq},

{eq}\begin{align} \ln \left(3x\right)&=\ln \left(\frac{1}{e^{\frac{1}{2}}}\right)\\ 3x&=\frac{1}{e^{\frac{1}{2}}}\\ x&=\frac{1}{3e^{\frac{1}{2}}} \end{align} {/eq}


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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